分类： ACM_HDU ACM_图论2013-10-27 13:37 226人阅读 评论(0) 收藏 举报
差分约束
题意:
给你一个N*M的矩阵C，问是否存在一个长度为N的数列a和长度为M的数列b使得所有C(i,j)*a(i)/b(j)在L , U范围内。
解题思路:
有两个式子 C(i,j)*a(i)/b(j) >= L , C(i,j)*a(i)/b(j) <= U ，两边均取对数得到 log(i) - log(j) <= log( U/C(i , j) ) 和 log(j) - log(i) <= log( C(i,j) / L) ，很显然的差分约束。。可惜用普通的队列版SPFA是会TLE的！网上说只需要入队sqrt(n)次就可以判断有负环了，不过没有具体的证明解释呀！所以正解应该是DFS版的也就是用栈的SPFA，因为要找负环，DFS找到一个负环后接下来相当于一直会绕着这个负环走个n+1次就跳出了，比BFS版判负环高效的多！
- /* **********************************************
- Author : JayYe
- Created Time: 2013-10-23 16:18:20
- File Name : JayYe.cpp
- *********************************************** */
- #include <stdio.h>
- #include <math.h>
- #include <string.h>
- #include <algorithm>
- using namespace std;
- const int maxn = 800 + 5;
- const int INF = 2000000000;
- struct Edge{
- int to, next;
- double w;
- }edge[maxn*maxn*2];
- int head[maxn], T[maxn], E, q[maxn*maxn];
- bool vis[maxn];
- void init(int n) {
- for(int i = 0;i <= n; i++)
- head[i] = -1;
- E = 0;
- }
- void newedge(int u, int to, double w) {
- edge[E].to = to;
- edge[E].w = w;
- edge[E].next = head[u];
- head[u] = E++;
- }
- double dis[maxn];
- bool SPFA(int n) {
- for(int i = 0;i <= n; i++) {
- dis[i] = INF; vis[i] = T[i] = 0;
- }
- T[0] = 1;
- dis[0] = 0;
- vis[0] = 1;
- int st = 0, ed = 0;
- q[++ed] = 0;
- while(ed) {
- int u = q[ed--];
- //if(st > n*n) return false;
- vis[u] = 0;
- for(int i = head[u];i != -1;i = edge[i].next) {
- int to = edge[i].to;
- double w = edge[i].w;
- if(dis[to] > dis[u] + w) {
- dis[to] = dis[u] + w;
- if(!vis[to]) {
- vis[to] = 1;
- T[to]++;
- if(T[to] > n + 1 ) return true;
- q[++ed] = to;
- }
- }
- }
- }
- return false;
- }
- int main() {
- int n, m, L, U, c;
- while(scanf("%d%d%d%d", &n, &m, &L, &U) != -1) {
- init(n+m);
- for(int i = 1;i <= n; i++) {
- for(int j = 1;j <= m; j++) {
- scanf("%d", &c);
- newedge(j + n, i, log((double)U) - log((double)c));
- newedge(i, j + n, log((double)c) - log((double)L));
- }
- }
- for(int i = 1;i <= n+m; i++) newedge(0, i, 0);
- if(!SPFA(n + m)) puts("YES");
- else puts("NO");
- }
- return 0;
- }
